Question

please show a step by step of a,b, and c. Thank you!

Consider a function f : R R such that f'(x) = x²(x - 1)(x - 2). Find all stationary points of this functie and determine their types. Find the globally optimal solution to max 13 - 2 s.t. -1<r<2 Find all local solutions to max 23 - 3x2 + 4x - 1 s.t. -2<3<4

Answer 1

- (a) Given f(0) = x²(x-1) (x-2) Now = = f'(X) = 0 x2(x-1) (X-2) = 0 x=0, (0-1) = 0, 0-2) = 0 = x=0,0, x=1 , x=2 =) Now Stationary points of f are [0, 1, and 2 . f'(o) = ¢ too= d[x²(x-1)(x-2)] de Felipe (x?ex?-3x +2)]
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of (x)= d [xl-3x3 + 2x2] = 4x3 - 9x² + 4x Rod Now at x=0 f" (o)=4.03–9.02 +4.0 o is point of Inflection. at x=1 f" (1) = 4.3 -9.12 +4.1 = u raty = + <o = x=1 is point of minima. at x=20 f" (2) - U.23 -9.22+4.2 =4.8 -9.4 +8 L = 32-36 + 8 = 40-36= uto = x=2 is point of maxima.

(b) we have Now a x3-X f'(x)=d = fex) [say] 0c3_x) do f'(x) = 0 => 3x2-1=0 => 3r? EiX21 V3 Since we want Globally optimall Solution So we need to check value ad not only Stationam points it but also boundary points 4 82 Now f(-) = (-1)}-(-1) = 1 +1 0 f(-) =()3- (- 2) = 353 & f(t) = ()? 13 f(2)= 23-2 = 8-2 = 6 on 18x62 Global optimal value of ad x3 x is 6-

Sol Let too = x3 - 3x2 + 4x - 1 f'(x) = 3x² - 6x + 4 f'(x) = 0 73x2 - 6x tuzo 9 3 = -(-6) + 16632-4,3.4 = 6+ / 36 - 48 Here discriminant 36-48 = 5-12 1) roots of 1 3x2_6x+u are o . So f(x) does not have any Stationary points du we need to check values only at boundars points x=-2 & 4 Now f(-2) = €2)3 – 3(+2)2 +4(-2) - 1 --8-12-8-1 = -29 f (4) = 43 -3.42 +4.4-1 = 64-u8 +167 -310